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estudierBest ResponseYou've already chosen the best response.0
final answer to what?
 one year ago

aroubBest ResponseYou've already chosen the best response.1
\[\sqrt[10]{x^4} \times \sqrt[10]{x^3} \times \sqrt[10]{x^3}\]
 one year ago

estudierBest ResponseYou've already chosen the best response.0
Can't even see that, is it 10 th root?
 one year ago

akash_809Best ResponseYou've already chosen the best response.0
see moderator has warned me now...whats the way to delet ur account...can somebody give the link...i m going frm here
 one year ago

akash_809Best ResponseYou've already chosen the best response.0
can u pls give me the link...i want to delet my account..i can't find it anywhere
 one year ago

aroubBest ResponseYou've already chosen the best response.1
I don't know.. You have to contact a moderator
 one year ago

akash_809Best ResponseYou've already chosen the best response.0
cmon moderator show up now...
 one year ago

phiBest ResponseYou've already chosen the best response.0
The only reason I can think of is they want to make sure you get a real and postiive (as opposed to complex) root. There will be 10 different roots.
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
@akash_809 you cannot delete OS accounts
 one year ago

aroubBest ResponseYou've already chosen the best response.1
But you can write it without absolute value, right? Because (or this is what I've been taught) there's x^3 in the radicand so it's definitely positive..
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
the simple reason is that\[x^{10}=(x)^{10}\]so \[\large\sqrt[10]{x^{10}}\]will give you the same answer whether x is negative or positive
 one year ago

akash_809Best ResponseYou've already chosen the best response.0
what...thats cynical...open study community is surely going nowher with all these warnings and suspensions...and to top that off , you even cant delete your account...what is this open study team please response...i wnat to know how can i delete my account
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
absolute value of x is actually defined in 2 ways: 1) distance from the point x=0 on the number line 2) \(x=\sqrt{x^2}\)
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
so since\[\sqrt{(x)^2}=\sqrt{x^2}=x\]then\[\sqrt[10]{x^{10}}=\sqrt[10]{(x)^{10}}=x\](because x to any even power is positive, whether x>0 or x<0)
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
does that help you @aroub ?
 one year ago

aroubBest ResponseYou've already chosen the best response.1
In this case I totally understand why. But in the question itself there's an odd exponent in the radicand so it's definitely positive right? So then no need for an absolute value..
 one year ago

akash_809Best ResponseYou've already chosen the best response.0
@TuringTest root of negative number is not defined...it will result in iota value...so x wont be the answer i think
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
not necessarily, let's assume x<0...
 one year ago

aroubBest ResponseYou've already chosen the best response.1
If you assumed that, the question will be wrong.
 one year ago

akash_809Best ResponseYou've already chosen the best response.0
yes it is...\[lets take an example......\sqrt{2} is \not defined unless you are taking imaginary numbers into account then u wl get +1.414 i\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
the root of a negative number is not undefined, it is imaginary...\[\sqrt[10]{x^4} \times \sqrt[10]{x^3} \times \sqrt[10]{x^3}=\sqrt[10]{x^4}\times\sqrt[10]{x^3\cdot x^3}\]
 one year ago

akash_809Best ResponseYou've already chosen the best response.0
sry reply got messed up i was just saying that square root of 2 is not defined unless you consider imaginary numbers as wel..
 one year ago

aroubBest ResponseYou've already chosen the best response.1
Put imaginary numbers on a side for now..
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
imaginary numbers are defined... now if x<0, then x^3<0 that means that (x^3)(x^3)>0 which is what we would expect given that x^6>0 since all numbers to a positive exponent are >0
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
another way to see it, let's assume x>0\[\sqrt[10]{x^4} \times \sqrt[10]{x^3} \times \sqrt[10]{x^3}=\sqrt[10]{x^4}\times\sqrt[10]{x^3\cdot x^3}=\sqrt[10]{x^{10}}=x\]now we do it with the negative of x\[\sqrt[10]{(x)^4} \times \sqrt[10]{(x)^3} \times \sqrt[10]{(x)^3}=\sqrt[10]{(x)^4}\times\sqrt[10]{(x)^3\cdot (x)^3}\]\[=\sqrt[10]{(x)^{10}}=\sqrt[10]{x^{10}}=x\]so you see, after all is said and done we do not know whether the x we had at the beginning was positive or negative, hence we can only conclude that the answer is x
 one year ago

aroubBest ResponseYou've already chosen the best response.1
Okay take this as another example: \[4\sqrt[4]{5a^2}\times 8\sqrt[4]{125a}\times 2\sqrt[4]{a}\] Here, a does not need an absolute value. Why? You can assume that a>0
 one year ago

akash_809Best ResponseYou've already chosen the best response.0
yes you would have to assume a >0
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
let's just look at powers of a...
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
if a>0\[\sqrt[4]{a^2}\times\sqrt[4]a\times\sqrt[4]a=\sqrt[4]{a^4}=a=a\](since we already knew a>0 now with the negative of a...
 one year ago

akash_809Best ResponseYou've already chosen the best response.0
yea i m waiting for that negative values of a...what would it give
 one year ago

aroubBest ResponseYou've already chosen the best response.1
Aoh, so you have to try the negative and the positive?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
\[\sqrt[4]{(a)^2}\times\sqrt[4]{a}\times\sqrt[4]{a}=\sqrt[4]{a^4}=a\neq a\]because a>0, but we just assumed a<0 ! a contradiction, hence the assumption makes a difference
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
sorry, I mean we took the negative of a...
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
let me write that again, gotta be careful here
 one year ago

akash_809Best ResponseYou've already chosen the best response.0
that's the problem...the thing you did here is called fallacy of maths...http://en.wikipedia.org/wiki/Mathematical_fallacy ....search invalid roots in this article
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
assume a>0\[\sqrt[4]{a^2}\times\sqrt[4]a\times\sqrt[4]a=\sqrt[4]{a^4}=a=a\]assume a<0, then\[\sqrt[4]{a^2}\times\sqrt[4]a\times\sqrt[4]a=\sqrt[4]{a^4}=a\neq a\]since we assumed a<0
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
@akash_809 yes but at least one of these numbers is positive because it is raised to an even power, so I have committed no fallacy
 one year ago

akash_809Best ResponseYou've already chosen the best response.0
did u went through the link......the proofs given like \[1=\sqrt{1}=\sqrt{(1)(1)}=\sqrt{1}\sqrt{1}=i.i=1 ....so u hv 1=1...and certainly \it is \not valid..\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
same thing with the earlier problem...
 one year ago

akash_809Best ResponseYou've already chosen the best response.0
and 1=1....surely not right...
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
yes, but\[\sqrt{(1)^2}\cdot\sqrt{1}=\sqrt1\cdot\sqrt{1}\] is not the same as\[\sqrt{1}\cdot\sqrt{1}\]
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
notice we have a^2 in the above, so we know *that* part is positive
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
this turned out to be more subtle than I bet you realized @aroub :)
 one year ago

aroubBest ResponseYou've already chosen the best response.1
lol, yeah true xD ALWAYS thought absolute values are so easy! And whenever someone asks me about absolute values I answer until this question came up .' This question is a bit confusing >.< Well, THANKS LOOAADSSSSSS :D
 one year ago
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